Jul 9, 1996 - constrained Harnack inequality for the curve shortening ow has the interesting ...... orem. 6. Linear trace Harnack inequality for the Ricci flow in.
Constrained and linear Harnack inequalities for parabolic equations Bennett Chow and Richard Hamilton July 9, 1996
1. Introduction Dierential Harnack inequalities for parabolic equations originated with the work of Li and Yau [LY] on the heat equation on a Riemannian manifold. Perhaps most importantly for problems in dierential geometry is that their technique relies only on the maximum principle. Dierential Harnack inequalities have now been proven for many geometric evolution equations using the maximum principle. In particular, the second author has proved Harnack inequalities for the Ricci ow [H1], [H2], the mean curvature ow [H3], and a matrix Harnack inequality for the heat equation [H4]. Similar Harnack inequalities have been proven by the rst author for the Gauss curvature ow [C1] and the Yamabe ow [C2]. Harnack inequalities have also been proven by Cao [Ca] for the Ricci-Kähler ow and by Andrews [A] for general curvature ows of hypersurfaces. In this paper we shall give extensions of the Harnack inequalities for the heat equation, Ricci ow on surfaces, and curve shortening ow, which we shall call constrained Harnack inequalities. We shall also generalize the trace Harnack inequality for the Ricci ow in all dimensions in another direction, which we shall call a linear Harnack inequality. In the case of the Ricci ow on surfaces, we shall actually obtain a constrained linear Harnack inequality. The constrained Harnack inequality for the heat equation was discovered by trying to extend the constrained gradient estimates for p > 1 to the case p = 1. Extending the constrained Harnack inequality to the Ricci ow on surfaces led to the discovery of the linear Harnack inequality, which generalizes to higher dimensions. The constrained Harnack inequality for the curve shortening ow has the interesting feature that the auxiliary equation is satis ed by a geometric quantity, namely the
support function. The underlying motivation of our work is to better understand how the Harnack inequality for the Ricci ow can be perturbed or extended, which is especially important in dimension three for studying singularities of the Ricci ow.
2. Constrained gradient estimate for the heat equation In this section we prove the following gradient estimate for a solution to the heat equation constrained by a supersolution. For simplicity, we consider the case of the unit ball in Euclidean space. Analogous results also hold on the entire Euclidean space and on Riemannian manifolds. Theorem 2.1. Suppose f is a solution to the heat equation and g is a supersolution @f = 4f and @g 4g @t @t on fx 2 R n : jxj 1g for t 0; and suppose
jf jp g
for some p > 1: Then we can nd a constant C (depending only on p and the dimension) so that 1 1 1 =p jrf j C g p + : (2.1)
t 1 , jxj
We rst note the following useful Lemma 2.2. Given any nonnegative functions P and Q and constants and ; we have
P
P ,1 2P , P 2Q Q Q +1 ,2 , ( , 1) PQ jrP j2 , ( + 1) QP +2 jrQj2 P ,1 +2 +1 hrP; rQi ; Q where 2 = @t@ , 4 is the heat operator. In particular, if + 1 1; then P P ,1 2 Q Q 2P , QP +1 2Q:
2 Q
=
2
We shall give two proofs of Theorem 2.1. Proof 1. Case 1: p 2: We compute
2 gp 2
, f2
2 2
, p p , 1 g p ,2 jrgj2 + 2 jrf j2 2 jrf j2 ;
since p 2: Also we have
2
2 jrf j 0;
from the usual computation
2 jrf j = , jrf j,3 jrf j2 jrrf j2 , jrf rrf j2 : Applying Lemma 2.2 with P = jrf j ; Q = g p , f 2; = 2; and = 1; we have 2
!
!2
f j2 ,2 jrf j2 : 2 jr gp , f2 gp , f2 2
2
We now apply the usual barrier argument. Let
'(x; t) = 1t + ,
1
1 , jxj2
2 :
We compute that 2 2' = , t12 , , 4n 2 3 , , 24 jxj 24 ,C'2 ; 1 , jxj 1 , jxj where C is a constant depending only on n: Hence, choosing C large enough, we
have
2 (C ') ,2 (C ')2 : Since ' ! 1 as either t ! 0 or jxj ! 1; we may apply the maximum principle
to conclude
jrf j2 C ' C p1 + 1 2 ; t 1 , jxj gp , f2 2
which proves our result (with C independent of the value of p 2) when we drop
f 2 from the left-hand-side.
3
Case 2: 1 < p < 2: In this case let P = jrf j and Q = g , 2pf ,2 1 ; where and g are to be chosen later. We compute for 1 < p 2
2Q =
2
g p ,1 2
f2 2 jr g j +1 p p
jrf j2 + 2 , 1 2 2
p
f
g
2
,4 p , 1 p hrf; rgi g Cp ,1 jrf j2 ; g
2
2
where C > 0 is a positive constant depending only on p: By Lemma 2.2, for + 1 1; we have
0 1 B jrf j CC ,C jrf j +2 : 2B + p @ A f f g , g p2 ,2 1
2
g , g p2 ,1 2
1 and Hence, if we assume in addition that = p (e.g., we may take = p,1 p ,) then = p,1
0 1 0 11+ p p p B C B CC : jr f j jr f j B C B 2@ , C A @ A f f 2
g , g 2p ,2 1
g , g p2 ,2 1
In our barrier argument, now let = p 2 and (x; t) =
so that
1
1 +, t 1 , jxj2 2 1
2 ,C
By the maximum principle, we conclude
p jr f j C t 1p g , g pf , 2
2
2 1
4
1+ :
1
!
+, p ; 1 , jxj2
1 where C depends only on n, p, and . The theorem follows from taking = p,1 say. Note that the constant C in (2.1) gets worse as p tends to 1. It is also interesting to observe that the quantities we consider in the two cases coincide when p = 2: In the second proof we do not need to separate the argument into two cases. However, we then have the di culty of handling the gradient term in the evolution equation. Proof 2. Let
P H=Q
1 with P = jrf j ; Q = g p , fp12 ; and 0 < 1p < 1: Recall that 2P 0: We also g compute
2 jr g j 2g c 2, p 1
p
!
and
1
2 2 jr f j f 2 p1 ,c p1 g g
so that
!
2 jrf j2 jr g j 2Q c 2, p + p :
Now for a general quotient
P
2 Q so that
g
g
g
1
1
P 2P P 2Q r Q =2 r + , ; Q
Q
Q
Q2
!
2 2 2H 2 rQQ rH , c jrgg2 j + jrfp j H;
g2
where we used H jrp1f j : To estimate the rst term on the RHS, we compute g
!
2 rQ = p g p + f p1 rgg , 2f r1p f g g
1
1
5
!
so that
jrQj C jr1,gpj + jf j jrp f j ; g
g
1
1
using fp12 g p : We may assume without loss of generality, by replacing g by 2g g and noting that the desired estimate (2.1) only changes by a constant factor, that 1
jf jp 12 g:
(2.2)
!
jrQj2 C jrgj2 + jrf j2 ; Q2 g2 p g
2
where we used Q c g p (which follows from (2.2)) and jf j g p : From 1
2
1
rQ
1 jrH j2 jrQj2 ; r H + 2 H; Q H Q
we conclude
!
1 jrH j2
2 2 2H H + (C , c) jrgg2 j + jrf j H: gp 2
Taking small enough, we get 2 jr H j 2H C H , cH 3
for some C < 1 and c > 0: Choose the barrier function
1
1
: =B p + t 1 , jxj2 If B is large, we nd whence H as desired.
2 jr j 2 C , c3;
6
3. Constrained trace and matrix Harnack inequalities for the heat equation In this section we extend both Li and Yau's [LY] dierential trace Harnack inequality and the second author's [H4] dierential matrix Harnack inequality for the heat equation on a Riemannian manifold. First we state our result for a multiply-periodic function on R n ; or equivalently, for an n-torus.
Theorem 3.1. If f and g are two solutions to the heat equation on an n-torus (or any other compact at Riemannian manifold)
@f = 4f @t
and
@g = 4g; @t
(3.1)
with jf j < g; then we have the matrix inequality
@ 2 ln g + 1 > 1 @h @h ; @xi @xj 2t ij 1 , h2 @xi @xj
and in particular, the trace inequality
@ ln g , jr ln gj2 + n = 4 ln g + n > jrhj2 : @t 2t 2t 1 , h2
This result extends to the heat equation on compact Riemannian manifolds. It should also extend to Euclidean space and other noncompact Riemannian manifolds. Let (M; ) be a compact Riemannian manifold. We again consider two solutions f and g to the heat equation (3.1) with jf j < g: We shall prove the following constrained trace and matrix Harnack inequalities.
Theorem 3.2. If (M; ) has nonnegative Ricci curvature, then @ ln g , jr ln gj2 + n = 4 ln g + n > jrhj2 ; @t 2t 2t 1 , h2
where h = f=g:
Theorem 3.3. If (M; ) has nonnegative sectional curvature and parallel Ricci curvature, then rirj ln g + 21t ij > r1i,hrhj2h : 7
Remark. If f Cg; where C 2 (,1; 1) is a constant, then rh 0 and we recover
the Harnack inequalities in [LY] and [H4]. Furthermore, less sharp inequalities should hold depending on a lower bound for the Ricci curvature in the trace case and depending on a lower bound for the sectional curvature and a bound on the norm of the covariant derivative of the Ricci tensor in the matrix case. We expect that the results also hold on complete noncompact Riemannian manifolds under suitable hypotheses. All of the results above are derived from the following computation, where in the case of a at manifold such as an n-torus, the curvature terms drop out. Let
hrj h : Pij = rirj L , r1i, h2
We have
Lemma 3.4.
@ P = 4P + 2r Lr P + 2P P , R P , R P ij l l ij il lj il lj jl li @t ij 2hri hrl h 2hrj hrl h 2 rirl h + 1 , h2 r j rl h + 1 , h 2 + 1 , h2 r hr h +2Rkijl Pkl + 2Rkijl k l2 + 2Rkijl rk Lrl L 1,h , (riRjl + rj Ril , rl Rij ) rl L:
Proof. Let L = ln g. We compute that @ L = 4L + jrLj2 : @t
A straightforward computation yields the general result that if a function A satis es the equation where B is some function, then
@A = 4A + B; @t
@ r r A = 4r r A + 2R r r A , R r r A , R r r A i j kijl k l il j l jl i l @t i j , (riRjl + rj Ril , rl Rij ) rl A + rirj B: 8
Applying this result to A = L = ln g; we obtain
@ r r L = 4r r L + 2R r r L , R r r L , R r r L i j kijl k l il j l jl i l @t i j , (riRjl + rj Ril , rl Rij ) rl L +2ri rl L rj rl L + 2rl ri rj L rl L , 2Riljm rl Lrm L:
i hrj h Next we compute the evolution of the term r1, h2 : The evolution of the ratio h = f=g is given by
and its gradient satis es
@ h = 4h + 2 hrL; rhi ; @t
@ (rh) = r @ h = r (4h + 2 hrL; rhi) @t @t = 4rh + 2 hrrL; rhi + 2 hrL; rrhi , Rc(rh);
which implies
@ (r hr h) = 4 (r hr h) , 2r r hr r h + 2r r Lr hr h + 2r r Lr hr h i j i l j l i l l j j l l i @t i j +2rl Lrl (ri hrj h) , Ril rl hrj h , Rjl rl hri h: We also compute
@ ,1 , h2 = ,2h @ h = ,2h (4h + 2 hrL; rhi) @t @t
, , = 4 1 , h2 + 2 rL; r 1 , h2 + 2 jrhj2 :
Combining the above two equations, we obtain
@ ri hrj h = 4 rihrj h + 2r Lr rihrj h , rihrj h 2 jrhj2 l l @t 1 , h2 1 , h2 1 , h2 (1 , h2 )2 1 , 2ri rl hrj rl h + 2ri rl Lrl hrj h + 1 , h2 +2rj rl Lrl hri h , Ril rl hrj h , Rjl rl hri h , 4h 2 2 rl h [ri rl hrj h + rihrj rl h] (1 , h ) 2 jh , 8h ri hr jrhj2 : 3 2 (1 , h ) 9
Rearranging terms yields
@ rihrj h = 4 rihrj h + 2r Lr ri hrj h l l @t 1 , h2 1 , h2 1 , h2 2 2hri hrl h 2hrj hrl h , 1 , h2 rirl h + 1 , h2 rj rl h + 1 , h2 2r r Lr hr h + 2r r Lr hr h 1 i l l j j l l i + 2 , R r h r h , R 1,h il l j jl rl hri h , rihr2j h2 2 jrhj2 : (1 , h )
Now we let
hrj h : Pij = rirj L , r1i, h2
Combining our two previous computations, we have
@ P = 4P + 2r Lr P + 2P P , R P , R P ij l l ij il lj il lj jl li @t ij 2hri hrl h 2hrj hrl h 2 rirl h + 1 , h2 r j rl h + 1 , h 2 + 1 , h2 r hr h +2Rkijl Pkl + 2Rkijl k l2 + 2Rkijl rk Lrl L 1,h , (riRjl + rj Ril , rl Rij ) rl L;
and the proof of the lemma is complete.
Tracing the lemma, we obtain the following Corollary 3.5. If we de ne
P then
= g ij Pij = 4L ,
jrhj2 ; 1 , h2
@ P = 4P + 2 hrL; rP i + 2 rrL , rhrh 2 @t 1 , h2 2hrhrh + ,1 , h2 rrh 2 2 + (1 , h2 )3 1 +2Rc(rL; rL) + 2Rc(rh; rh): 1 , h2 10
We are now ready to prove the theorems. Proof of Theorem 3.2. Applying the elementary inequality
jaij j2 n1 (tr a)2
to the corollary and using the assumption that the Ricci curvature is nonnegative, we obtain Therefore
@ P 4P + 2 hrL; rP i + 2 P 2: @t n
@ P + n 4 P + n + 2 DrL; r P + n E @t 2t 2t 2t n n 2 P + 2t ; + P, n 2t
and the theorem follows from applying the maximum principle to this equation.
Proof of Theorem 3.3. By the lemma and the hypotheses that rRc 0 and K 0; we have @ P 4P + 2r Lr P + 2P P , R P , R P + 2R P : ij l l ij il lj il lj jl li kijl kl @t ij
Therefore
@ P + 1 g = 4 P + 1 g + 2r Lr P + 1 g ij ij l l ij @t ij 2t ij 2t ij 21t 1 +2 Pil , gil Pjl + gjl 2t 2t
,Ril Plj + 21t glj
, Rjl Pli + 21t gli ;
and by applying the maximum principle, we obtain
Pij + 21t gij > 0:
Proof of Theorem 3.1. This is just a special case of Theorem 3.3 since an n-torus is at.
11
4. Linear and constrained trace Harnack inequalities for the Ricci ow on surfaces In this section we generalize the second author's [H1] trace Harnack inequality for the Ricci ow on surfaces with positive scalar curvature. We shall prove a constrained linear trace Harnack inequality which has as a special case the linear trace Harnack inequality. Let g(t) be a solution to the Ricci ow on a surface M 2 with R > 0 :
@ g = ,R g : ij @t ij
The evolution of the scalar curvature is then given by
@R = 4R + R2: @t
The second author proved that if the initial metric has positive scalar curvature, then @ ln R , jr ln Rj2 + 1 = 4 ln R + R + 1 > 0: (4.1) @t t t We shall prove the following generalizations of this Harnack inequality. Let S and T be solutions to the linear equations
@S = 4S + RS @t @T = 4T + RT; @t
(4.2) (4.3)
where 4 and R are the Laplacian and scalar curvature of the metric moving under Ricci ow, with the property that initially
jT j < S: By the maximum principle, this inequality is preserved under the Ricci ow. We have the following constrained linear Harnack inequality. Theorem 4.1. If (M 2 ; g(t)) is a solution to the Ricci ow on a surface M 2 with R > 0, then for any solutions S and T to (4.2) and (4.3) with jT j < S;
@ ln S , jr ln S j2 + 1 = 4 ln S + R + 1 > jrhj2 ; @t t t 1 , h2 where h = T=S: 12
Taking T 0; we obtain the following linear Harnack inequality.
Corollary 4.2. If g(t) is a solution to the Ricci ow on a surface with R > 0; and S a solution to (4.2), then
Remark. Since
@ ln S , jr ln S j2 + 1 = 4 ln S + R + 1 > 0: @t t t @R = 4R + R2; @t
we may take S = R; and we then obtain the trace Harnack inequality for the Ricci ow on surfaces (4.1) as a special case.
Proof of Theorem 4.1. Let We rst compute
@ ln S , jr ln S j2 : Q = 4 ln S + R = @t
@Q = 4 @ ln S + @ 4 ln S + @R @t @t @t @t , 2 = 4 Q + jr ln S j + R 4 ln S + 4R + R2 = 4Q + 2 jrr ln S j2 + 2r ln S r 4 ln S + R jr ln S j2 +R 4 ln S + 4R + R2 2 1 = 4Q + 2 rr ln S + Rg + 2r ln S rQ 2 +R jr ln S , r ln Rj2 , RQ + R (4 ln R + R) :
Hence, we have
@Q 4Q + 2r ln S rQ + 2 rr ln S + 1 Rg 2 , RQ + R (4 ln R + R) : @t 2 Next we compute that h = T=S satis es @ h = 4h + 2 hr ln S; rhi : @t 13
The evolution of the gradient of h is given by
@ rh = 4rh + 2 hrr ln S; rhi + 2 hr ln S; rrhi , 1 Rrh; @t 2
which implies
@ jrhj2 = 4 jrhj2 , 2 jrrhj2 + 4 hrr ln S; rhrhi + 2 r ln S; r jrhj2 : @t
We also have
@ ,1 , h2 = 4 ,1 , h2 + 2 r ln S; r ,1 , h2 + 2 jrhj2 : @t Recall that, in general, if two functions F and G satisfy equations of the form @F = 4F + A @t @G = 4G + B; @t then
@ F = 4 F + 2 hrG; rF i , 2F jrGj2 + A , FB : @t G G G2 G3 G G2
From this we compute
@ jrhj2 @t 1 , h2
Now let We have
!
=
!
*
!+
2 2 4 1jr,hhj 2 + 2 r ln S; r 1jr,hhj 2 , , 2 2 3 2hrhrh + 1 , h2 rrh 2 (1 , h ) 4 2 jrhj4 + hrr ln S; rhrhi , : 1 , h2 (1 , h2 )2
2 2 P = Q , 1jr,hhj 2 = 4 ln S + R , 1jr,hhj 2 :
@ P 4P + 2r ln S rP + 2 rr ln S + 1 Rg 2 , RQ + R (4 ln R + R) @t 2 14
2hrhrh + ,1 , h2 rrh 2 3 (1 , h2 ) 4 , 1 ,4 h2 hrr ln S; rhrhi + 2 jrh2j 2 (1 , h )
+
=
Hence
2
2 1 r h r h 4P + 2r ln S rP + 2 rr ln S + 2 Rg , 1 , h2 2hrhrh + ,1 , h2 rrh 2 + R jrhj2 2 + 1 , h2 (1 , h2 )3 ,RP + R (4 ln R + R) 4P + 2r ln S rP + P 2 , RP + R (4 ln R + R) :
@ P+1 @t t
1
1
1
4 P + t + 2r ln S r P + t + P , t 1 1 ,R P + t + R 4 ln R + R + t :
1
P+t
The theorem now follows from using the trace Harnack inequality
4 ln R + R + 1t > 0;
while applying the maximum principle to the evolution equation for P + 1t .
5. Constrained matrix Harnack inequality for the Ricci ow on surfaces In [H2], the second author proved a matrix Harnack inequality for the Ricci ow on compact Riemannian manifolds with nonnegative curvature operator. When n = 2; the inequality says that if R > 0; then
rirj ln R + 21 Rgij + 21t gij > 0:
We shall prove the following extension to a constrained Harnack inequality, which generalizes Theorem 4.1 in the case where S = R: 15
Theorem 5.1. Let (M; g(t)) be a solution to the Ricci ow on surfaces with R > 0. If T is a solution to
with jT j < R; then where h = T=R:
Proof. Let
@T = 4T + RT @t
rirj ln R + 12 Rgij + 21t gij > r1i,hrhj2h ; Qij = rirj ln R + 21 Rgij :
Its evolution equation is given by
Lemma 5.2.
@ Q = 4Q + 2r ln Rr Q + 2Q Q + RQg , 3RQ : ij k k ij ik jk ij ij @t ij Proof. We rst recall the following general formula. If a function A satis es the equation @ A = 4A + B; @t then under the Ricci ow on surfaces, we have
@ r r A = 4r r A + r r B , 2R r r A , 1 4 A g : i j i j i j ij @t i j 2
(5.1)
Indeed, we compute
@ r r A = r r @ A , @ ,k r A i j @t i j @t @t ij k 1 = ri rj (rk rk A + B ) + g kl (ri Rgjl + rj Rgil , rl Rgij ) rk A 2 = ri rk rj rk A , ri (Rjp rpA) + ri rj B 1, + ri Rrj A + rj Rri A , rl Rrl A gij 2 16
so that
@ r r A , r r B , 1 ,r Rr A + r Rr A , r Rrl A g i j j i l ij @t i j 2 i j 1 = rk ri rj rk A , Rikjprp rk A , Rip rj rpA , ri (Rrj A) 2
= = =
rk rk ri rj A , rk (RikjprpA) , Rikjprprk A , Riprj rpA , 12 ri (Rrj A) 4rirj A , rk RikjprpA , 2Rikjprprk A , 21 Rrj riA , 12 ri (Rrj A) 4rirj A , 21 (rj RriA , gij rpRrpA) , R (ri rj A , gij rk rk A)
,Rrj riA , 12 riRrj A
and (5.1) follows. Now since
@ ln R = 4 ln R + jr ln Rj2 + R; @t
using above formula, we obtain
@ Q = @ r r ln R + 1 Rg @t ij @t i j 2 ij , 1 2 = 4Qij + ri rj jr ln Rj + R , 2R ri rj ln R , 4 ln R gij 2 = 4Qij + 2rk ri rj ln Rrk ln R , 2Rikjprp ln Rrk ln R + 2ri rk ln Rrj rk ln R +ri rj R , 2RQij + RQgij , = 4Qij + 2rk Qij rk ln R , rk Rrk ln R gij , R ri ln Rrj ln R , gij jr ln Rj2 1 1 +2 Qik , Rgik Qjk , Rgjk + ri rj R , 2RQij + RQgij 2 2 =
4Qij + 2rk Qij rk ln R + rirj R , Rri ln Rrj ln R + 21 R2gij +2Qik Qjk , 4RQij + RQgij
and the lemma follows.
Now we return to the proof of the theorem. Let h = T=R: Recall from the last section that we have the following formulas (taking S = R)
@h = 4h + 2r ln R rh; @t 17
and Hence
@ rh = 4rh + 2rr ln R rh + 2r ln R rrh , 1 Rrh; @t 2 @ ,1 , h2 = 4 ,1 , h2 + 2r ln R r ,1 , h2 + 2 jrhj2 : @t
@ (r hr h) = 4 (r hr h) , 2r r hr r h , Rr hr h + 2r ln R r (r hr h) i j k i k j i j k k i j @t i j +2rj rk ln R ri hrk h + 2ri rk ln R rj hrk h:
Combining the above two equations yields
@ rihrj h @t 1 , h2
and hence
=
4
r hr h i
j
2
,
2 r k (ri hrj h) rk 1 , h 2 2 2 1,h (1 , h ) r hr h r 2 i hrj h , 2 r 1 , h + 2rk ln R rk 1i, hj2 ,2 3 2 (1 , h ) 1 , 2 r r h r r h , R r h r h k i k j i j + 1 , h2 +2rj rk ln R ri hrk h + 2rirk ln R rj hrk h ,2 rihr2j h2 jrhj2 ; (1 , h )
+
@ rihrj h = 4 rihrj h + 2r ln R r rihrj h k k @t 1 , h2 1 , h2 1 , h2 2hri hrk h 2hrj hrk h 2 rk rj h + 1 , h2 , 1 , h2 rk rih + 1 , h2 1 + (2rj rk ln R ri hrk h + 2ri rk ln R rj hrk h) 1 , h2 ,2 ri hr2j h2 jrhj2 , R r1i,hrhj2h : (1 , h )
i hrj h Combining the evolution equations for Qij and , r1, h2 ; we obtain
@ Q , rihrj h @t ij 1 , h2 r r i hrj h i hrj h = 4 Qij , + 2rk ln Rrk Qij , 1 , h2 1 , h2 18
+RQgij , 3RQij + 2Qik Qjk 2 2hri hrk h 2hrj hrk h + rk rih + 1 , h2 rk rj h + 1 , h2 1 , h2 , 1 ,1 h2 (2rj rk ln R ri hrk h + 2rirk ln R rj hrk h) r hr h r hr h +2 i 2j 2 jrhj2 + R i j2 : 1,h (1 , h )
Simplifying by combining terms, we have
@ Q , rihrj h @t ij 1 , h2 r r i hrj h i hrj h = 4 Qij , + 2rk ln Rrk Qij , 1 , h2 1 , h2
!
2 r jr h j i hrj h g , 3R Qij , 1 , h2 +R Q , 1 , h2 ij jrhj2 r r i hrk h j hrk h +2 Qik , Qjk , 1 , h2 + R 1 , h2 gij 1 , h2 2hri hrk h 2hrj hrk h 2 rk rih + 1 , h2 rk rj h + 1 , h2 : + 1 , h2
i hrj h ij Let Pij = Qij , r1, h2 and P = g Pij ; we then have
@ P = 4P + 2r ln Rr P + RPg , 3RP + 2P P + R jrhj2 g ij k k ij ij ij ik jk @t ij 1, h2 ij 2 2hri hrk h 2hrj hrk h + r r : k ri h + k rj h + 2 2 1,h 1,h 1 , h2
Hence and
@ P 4P + 2r ln Rr P + RP g , 3RP + 2P P ; ij k k ij ij ij ik jk @t ij
@ P + 1 g 4 P + 1 g + 2r ln Rr P + 1 g + R P + 1 g ij k k ij ij @t ij 2t ij 2t ij 2t ij t 1 1 1 ,3R Pij + 2t gij + 2 Pik , 2t gik Pjk + 2t gjk : 19
Applying the maximum principle to the evolution equation above yields the theorem.
6. Linear trace Harnack inequality for the Ricci ow in higher dimensions In this section we extend the linear trace Harnack inequality for the Ricci ow on surfaces (Corollary 4.2) to higher dimensions.
Theorem 6.1. Let (M; g(t)) be a solution to the Ricci ow @ g = ,2R ; ij @t ij
and let h be a symmetric 2-tensor satisfying
@ h = 4h + 2R h , R h , R h : ij pijq pq iq jq jq iq @t ij
(6.1)
Assume that the metric initially has nonnegative curvature operator, which is preserved under the ow. If in addition, hij > 0 initially, then hij > 0 as long as the solution exists, and for any vector eld V we have
Remarks.
> 0: div (div (h)) + Rij hij + 2div (h) V + hij ViVj + H 2t
1. The Ricci tensor Rij is a solution to (6.1). Thus as a special case, we have
4R + 2 jRcj2 + 2rR V + 2Rij ViVj + Rt > 0;
which is the trace Harnack inequality for the Ricci ow. 2. If f is a solution to the heat equation
@f = 4f; @t
where the laplacian is with respect to the metric evolving under the Ricci ow, then the hessian of f
hij = rirj f 20
is a solution to (6.1). In particular, if f is initially convex on a complete solution to the Ricci ow with nonnegative curvature operator on a noncompact manifold, and provided we are able to apply the maximum principle (we have not investigated when this is the case,) then
44 f + 12 rR rf +2Rij rirj f +2 (ri 4 f + Riprpf ) Vi + rirj fViVj + 42tf > 0: Note that here we used
div (rrf )i = rj rirj f = ri 4 f + Riprpf: 3. When n = 2; the symmetric 2-tensor
hij = u gij ; which is conformal to the metric, is a solution to (6.1) provided u satis es the equation
@u = 4u + Ru: @t
Hence Theorem 6.1 generalizes Corollary 4.2 to higher dimensions. 4. The rst author and S.-C. Chu [CC1] have given a geometric interpretation of the second author's Harnack inequality for the Ricci ow, where they show that the Harnack quadratic is the curvature of a torsion-free connection on space-time compatible with a degenerate metric on the cotangent bundle of space-time. Similarly, it can be shown that the linear trace Harnack inequality may be approached from this geometric viewpoint (see [CC2].) In particular, the linear trace Harnack inequality comes from linearizing the Ricci ow in a suitable way.
Proof of Theorem 6.1. We rst compute the evolution of the divergence of h : @ div (h) = @ ,gjk r h k ij i @t @t = g jk rk (4hij + 2Rpijq hpq , Riq hjq , Rjq hiq ) + 2Rjk rk hij @ @ p jk jk ,g @t ,ki hpj , g @t ,pkj hip = rj 4 hij + 2rj Rpijq hpq , rj Riq hjq , rj Rjq hiq +2Rpijq rj hpq , Riq rj hjq , Rjq rj hiq +2Rjk rk hij + (rj Rip + ri Rjp , rp Rij ) hpj : 21
Now
rj 4 hij
= = =
rj rprphij = rprj rphij , Rjpiq rphqj rprprj hij , rp (Rjpiq hqj , Rpq hiq ) , Rjpiq rphqj 4div (h)i , rpRjpiq hqj + rpRpq hiq , 2Rjpiq rphqj + Rpq rphiq ;
which implies
@ div (h) = 4div (h) , r R h + r R h , 2R r h + R r h p jpiq qj p pq iq jpiq p qj pq p iq i i @t +2rj Rpijq hpq , rj Riq hjq , rj Rjq hiq +2Rpijq rj hpq , Riq rj hjq , Rjq rj hiq +2Rjk rk hij + (rj Rip + ri Rjp , rp Rij ) hpj = 4div (h)i , 2rp Riq hpq + 2ri Rpq hpq , Riq div (h)q + 2Rjk rk hij : Hence we have the following evolution equation for div(h) @ div (h) = 4div (h) +2h r R , 2h r R +2R r h , R div (h) : (6.2) pq i pq pq p qi pq p qi qi i i q @t Second, we compute the evolution equation for the divergence squared of h : @ div (div (h)) = @ ,gij r div (h) i @t @t j @ @ ij r div (h) ij div (h)i + g = g rj j i @t @t = ri 4 div (h)i + 2hpq ri ri Rpq , 2hpq ri rp Rqi +2Rpq ri rp hqi , Rqi ri div (h)q +2ri hpq ri Rpq , 2ri hpq rp Rqi +2ri Rpq rp hqi , ri Rqi div (h)q + 2Rij rj div (h)i : Now
ri 4 div (h)i
= = =
rirprpdiv (h)i = rpr irpdiv (h)i rprpri div (h)i + rp Rpq div (h)q 4div (div (h)) + 21 rq Rdiv (h)q + Rpq rpdiv (h)q ; 22
which implies
@ div (div (h)) = 4div (div (h)) + 2r h r R , 2r h r R + 2r R r h i pq i pq i pq p qi i pq p qi @t +2hpq 4 Rpq , 2hpq ri rp Rqi + 2Rpq ri rp hqi + 2Rij rj div (h)i : Thus we have the following evolution equation for div(div (h)) :
@ div (div (h)) = 4div (div (h)) + 2r h r R i pq i pq @t 1 +2hpq 4Rpq , rp rq R + 4Rij rj div (h)i : 2 Our linear Harnack quadratic is de ned by
Z = div (div (h)) + Rij hij + 2div (h) V + hij ViVj ; where V is the vector eld minimizing Z; that is div (h)i + hij Vj = 0; or equivalently We compute
(6.3)
,
V j = , h,1 ji div (h)i :
@Z = 4div (div (h)) + 2r h r R + 2h 4R , 1 r r R + 4R r div (h) i pq i pq pq pq ij j i @t 2 p q +hij (4Rij + 2Rpijq Rpq , 2Riq Rjq ) + Rij (4hij + 2Rpijq hpq , Riq hjq , Rjq hiq ) +4Rkl Rik hil +2 4div (h)i + 2hpq ri Rpq , 2hpq rpRqi + 2Rpq rp hqi , Rqi div (h)q Vi @V + (4h + 2R h , R h , R h ) V V + 2h @Vi V ; +2div (h) ij pijq pq iq jq jq iq i j ij @t @t j
which we rewrite as a heat equation:
@Z = 4Z + 2h 4R , 1 r r R + 4R r div (h) + 4R R h pq pq ij j kl ik il i @t 2 p q +hij (2Rpijq Rpq , 2Riq Rjq ) + Rij (2Rpijq hpq , Riq hjq , Rjq hiq ) 23
+2 2hpq ri Rpq , 2hpq rp Rqi + 2Rpq rphqi , Rqi div (h)q Vi +2div (h)
@V
, 4V + (2Rpijq hpq , Riq hjq , Rjq hiq ) ViVj @t @V i +2hij @t , 4Vi Vj , 4rdiv (h) rV , 4rhij rViVj ,2hij rVirVj :
Simplifying and combining terms, we obtain
@Z = 4Z + 2h 4Rpq , 21 rprq R + 2RkpqlRkl , Rpr Rqr pq +2 (ri Rpq , rp Rqi ) Vi + Rpijq Vi Vj @t +4Rij rj div (h)i + 2hij Riq Rjq +2 2Rpq rphqi , Rqi div (h)q Vi + (,Riq hjq , Rjq hiq ) Vi Vj @V i +2 (div (h)i + hij Vj ) @t , 4Vi ,4rdiv (h) rV , 4rhij rViVj , 2hij rVirVj :
Using (6.3) and the following equation obtained by dierentiating it, rk div (h)i + Vj rk hij + hij rk Vj = 0; we obtain
(6.4)
@Z = 4Z + 2h 4Rpq , 21 rprq R + 2RkpqlRkl , Rpr Rqr pq +2 (ri Rpq , rp Rqi ) Vi + Rpijq Vi Vj @t +2hij (rk Vi , Rki ) (rk Vj , Rkj ) : Since the trace H of h satis es @H = 4H + 2R h ; pq pq @t we have
@ Z+H @t 2t
H + 2h 4Rpq , 1 rprq R + 2RkpqlRkl , Rpr Rqr = 4 Z+ pq +2 (r R 2, r R ) V + R V V + 1 R i pq p qi i pijq i j 2t pq 2t 1 1 +2hij rk Vi , Rki , gki rk Vj , Rkj , gki 2t 2t 2
1 + hij rj Vi , Rji , gji : t 2t
24
Since
Z = div (div (h)) + Rij hij + 2div (h) V + hij ViVj = div (div (h)) + Rij hij + div (h) V; we have
hij rj Vi , Rji , 21t gji
1 = ,ri div (h)i , Vj ri hij , hij Rji + gji 2t = =
and hence
,div (div (h)) , div (h) V , hij Rij , H 2t ,Z , H ; 2t
4R , 1 r r R + 2R R , R R @ Z + H = 4 Z + H + 2h pq p q kpql kl pr qr pq +2 (r R 2, r R ) V + R V V + 1 R @t 2t 2t i pq p qi i 1 pijq i j 2t pq 1 (6.5) +2hij rk Vi , Rki , gki rk Vj , Rkj , gki 2t 2t H 2 , t Z + 2t :
To analyze the second term on the right-hand-side, recall that the matrix Harnack inequality for the Ricci ow [H2] says that if the initial metric has nonnegative curvature operator, then for any 1-form Wi and any 2-form Uij
Y = Mpq WpWq , 2PqrpUqr Wp + RijklUij Ulk > 0; where and Setting
Mpq = 4Rpq , 21 rprq R + 2RkpqlRkl , Rpr Rqr + 21t Rpq Pqrp = rq Rrp , rr Rqp: Uij = 12 (WiVj , Wj Vi) ; 25
we have
Y = Mpq WpWq , Pqrp (Wq Vr , Wr Vq ) Wp + RijklWiVj Vk Wl = (Mpq , 2PqrpVr + Rpjkq Vj Vk ) Wp Wq 4R , 1 r r R + 2R R , R R + 1 R pq 2 p q kpql kl pr qr 2t pq W W 0: = p q ,2 (r R , r R ) V + R V V q rp
r qp
r
pjkq j k
Hence the second term on the right-hand-side is positive: 2hpq (Mpq , 2PqrpVr + Rpjkq Vj Vk ) 0:
@ Z+H 4 Z+H ,2 Z+H : @t 2t 2t t 2t
Thus
Applying the maximum principle, we have
H Z+H = div (div (h)) + Rij hij + 2div (h) V + hij Vi Vj + 2t 2t , H ij = div (div (h)) + Rij hij , h,1 div (h)i div (h)j + > 0: (6.6) 2t
This equation holds for
,
V j = , h,1 ji div (h)i ; however, since this V minimizes Z; equation (6.6) holds for all V; and the theorem
is proved.
7. Constrained Harnack inequality for the curve shortening ow We consider the curve shortening ow on a convex curve
@X = ,kN: @t
Here X : S 1 ! R 2 parametrizes a convex curve, k is the curvature, and N is the unit outward normal. The Harnack inequality for the curve shortening ow [H3] says that
@ 2 ln k + k2 + 1 > 0: @s2 2t 26
We shall prove a constrained Harnack inequality for the curve shortening ow which improves the above inequality. Recall that
@k = @ 2 k + k3: @t @s2
Let u be a solution to the equation
@u = @ 2 u + k2 u , 2k: @t @s2
The reason for considering this equation is that the support function hX; N i is a solution of the above equation (although the constrained Harnack inequality also works without the ,2k term.) Lemma 7.1. Under the curve shortening ow, the evolution of the support function is given by
Proof. Recall that
We compute
@ hX; N i = @ 2 hX; N i + k2 hX; N i , 2k: @t @s2 @N = @k T @t @s @N = kT @s @T = ,kN: @s
@ hX; N i = @ X; N + X; @ N = ,k + @k hX; T i : @t @t @t @s
On the other hand and
@ hX; N i = @ X; N + X; @ N = k hX; T i @s @s @s
@ 2 hX; N i = @ (k hX; T i) = @k hX; T i + k , k2 hX; N i : @s2 @s @s
The lemma follows.
Our constrained Harnack inequality for the curve shortening ow is as follows. 27
Theorem 7.2. If X : S 1 [0; T ) ! R 2 is a solution to the curve shortening ow with k > 0; and u : S 1 [0; T ) ! R is a solution to the equation with juj < k initially, then
@u = @ 2 u + k2 u , 2k; @t @s2
,
2 @ 2 L + k2 + 1 > @h @s ; @s2 2t 1 , h2 where h = uk +2t: In particular, if jhX; N ij < k initially, then the above inequality i is true for h = hX;N k + 2t:
Proof. We rst compute
@h = @ 2 h + 2 @L @h ; @t @s2 @s @s where L = ln k: Assume by rescaling that jhj < 1 initially, by the maximum principle, we have
jhj < 1;
as long as the solution exists. We compute that
@ @h 2 = @ 2 @h 2 , @ 2 h 2 + 4 @ 2 L @h 2 @t @s @s2 @s @s2 @s2 @s 2 2 @L @ @h @h 2 +2 @s @s @s + 2k @s
and
@ ,1 , h2 = @ 2 ,1 , h2 + 2 @L @ ,1 , h2 + 2 @h 2 : @t @s2 @s @s @s
From this we obtain
,
,
,
, !2
2 2 @h 2 2 @h 2 2h h @h @ @h @ @L @ 2 @ @s @s @s @s @t 1 , h2 = @s2 1 , h2 + 2 @s @s 1 , h2 , 1 , h2 @s2 + 2 1 , h2 , 2 , @h 4 @ 2 L , @h 2 k2 @h 2 @s @s @s +4 +2 , 2 @s 2 2 : 2 2 1,h 1,h (1 , h )
28
Recall that the Harnack quantity is and it satis es the equation
2 Q = @@sL2 + k2 ;
@Q = @ 2 Q + 2 @L @Q + 2Q2 : @t @s2 @s @s
, @h 2 , @h 2 @2 L 2 @s = + k , @s : P =Q,
Now de ne
1 , h2
We compute
@s2
1 , h2
, !2
2 @P = @ 2 P + 2 @L @P + 2 @ 2 L + k2 2 + 2 @ 2 h + 2 h @h @s @t @s2 @s @s @s2 1 , h2 @s2 1 , h2 , 2 !2 k2 , @h 2 , @h 4 2h h @h 2 @ @s @s + +2 , 2 1 , h2 + 2 @s 2 2 ; 1 , h2 @s2 1 , h2 (1 , h )
which implies
, !2
2 @P = @ 2 P + 2 @L @P + 2 @ 2 L + k2 , @h @s @t @s2 @s @s @s2 1 , h2 , 2 !2 k2 , @h 2 2h h @h 2 @ + + 2 @s 2 + 2 @s 2 : 1 , h2 @s2 1,h 1,h
Hence
@P @ 2 P + 2 @L @P + 2P 2; @t @s2 @s @s
,
and by applying the maximum principle to the equation for @t@ P + 21t , we obtain the theorem.
References [A]
Andrews, B. (1994) Harnack inequalities for evolving hypersurfaces, Math. Zeit. 217, 179-197. 29
[Ca] Cao, H.D. (1993) On Harnack's inequalities for the Kähler-Ricci ow, Invent. Math. 109, 247-263. [C1] Chow, B. (1991) On Harnack's inequality and entropy for the Gaussian curvature ow, Comm. Pure Appl. Math. 44, 469-483. [C2] Chow, B. (1992) The Yamabe ow on locally conformally at manifolds with positive Ricci curvature, Comm. Pure Appl. Math. 45, 1003-1014. [CC1] Chow, B. and Chu, S.C. (1995) A geometric interpretation of Hamilton's Harnack inequality for the Ricci ow, Math. Research Letters 2, 701-718. [CC2] Chow, B. and Chu, S.C. (1996) A geometric approach to the linear trace Harnack inequality for the Ricci ow, preprint. [H1] Hamilton, R.S. (1988) The Ricci ow on surfaces, Contemp. Math. 71, 237-262, Amer. Math. Soc., Providence, RI. [H2] Hamilton, R.S. (1993) The Harnack estimate for the Ricci ow, J. Di. Geom. 37, 225-243. [H3] Hamilton, R.S. (1995) The Harnack estimate for the mean curvature ow, J. Di. Geom. 41, 215-226. [H4] Hamilton, R.S. (1993) A matrix Harnack estimate for the heat equation, Comm. Anal. Geom. 1, 113-126. [LY] Li, P. and Yau, S.T. (1986) On the parabolic kernel of the Schrödinger operator, Acta Math. 156, 153-201. University of Minnesota University of California, San Diego
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